Question

9)A skier starts from rest from the top of a 40 m high slope which makes 40 degrees with the ground. Coefficient of friction is 0.1 What is the velocity of the skier at the bottom of the ramp?

Answer 1

The velocity of the skier at the bottom of the ramp is approximately 26.288 meters per second.

Step-by-step explanation:

We can determine the final velocity of the skier at the bottom of the ramp by Principle of Energy Conservation and Work-Energy Theorem, whose model is:

`U_(g,1)+K_(1) = U_(g,2)+K_(2)+W_(disp)` (1)

Where:

`U_(g,1)`,
`U_(g,2)` - Initial and final gravitational potential energy, measured in joules.

`K_(1)`,
`K_(2)` - Initial and final translational kinetic energy, measured in joules.

`W_(disp)` - Work dissipated by friction, measured in joules.

By definitions of gravitational potential and translational kinetic energy and work, we expand and simplify the model:

`m\cdot g \cdot (z_(1)-z_(2))+(1)/(2)\cdot m \cdot (v_(1)^(2)-v_(2)^(2)) =\mu_(k)\cdot N\cdot \Delta s` (2)

Where:

`m` - Mass, measured in kilograms.

`g` - Gravitational acceleration, measured in meters per square second.

`z_(1)`,
`z_(2)` - Initial and final heights of the skier, measured in meters.

`N` - Normal force from the incline on the skier, measured in newtons.

`\Delta s` - Distance covered by the skier, measured in meters.

`\mu_(k)` - Kinetic coefficient of friction, dimensionless.

The normal force exerted on the skier and the covered distance are, respectively:

`N = m\cdot g\cdot \cos \theta` (3)

`\Delta s = (z_(1)-z_(2))/(\sin \theta)` (4)

Where
`\theta` is the angle of the incline above the horizontal, measured in sexagesimal degrees.

By applying (3) and (4) in (2), we get that:

`m\cdot g \cdot (z_(1)-z_(2))+(1)/(2)\cdot m\cdot (v_(1)^(2)-v_(2)^(2)) = \mu_(k)\cdot m\cdot g \cdot \cos \theta \cdot \left((z_(1)-z_(2))/(\sin \theta) \right)`

`g\cdot (z_(1)-z_(2)) +(1)/(2)\cdot (v_(1)^(2)-v_(2)^(2))= \mu_(k)\cdot g \cdot \left((z_(1)-z_(2))/(\tan \theta) \right)` (5)

Then, we clear the velocity of the skier at the bottom of the ramp is: (
`v_(1) = 0\,(m)/(s)`,
`\mu_(k) = 0.1`,
`\theta = 40^(\circ)`,
`g = 9.807\,(m)/(s^(2))`,
`z_(1)-z_(2) = 40\,m`)

`\left[(\mu_(k))/(\tan \theta)-1 \right]\cdot g\cdot (z_(1)-z_(2)) = (1)/(2)\cdot (v_(1)^(2)-v_(2)^(2))`

`2\cdot \left[(\mu_(k))/(\tan \theta)-1 \right]\cdot g\cdot (z_(1)-z_(2)) = v_(1)^(2)-v_(2)^(2)`

`v_(2) = \sqrt{v_(1)^(2)-2\cdot \left[(\mu_(k))/(\tan \theta)-1 \right]\cdot g\cdot (z_(1)-z_(2))}` (6)

`v_(2) = \sqrt{\left(0\,(m)/(s) \right)^(2)-2\cdot \left((0.1)/(\tan 40^(\circ)) -1\right)\cdot \left(9.807\,(m)/(s^(2)) \right)\cdot (40\,m)}`

`v_(2) \approx 26.288\,(m)/(s)`

The velocity of the skier at the bottom of the ramp is approximately 26.288 meters per second.